Abstract Algebra: Groups and Subgroups

[taylor81792]

The problem says: Suppose that * is an associative binary operation on a set S.
Let H= {a ε S l a * x = x * a for all x ε s}. Show that H is closed under *. ( We think of H as consisting of all elements of S that commute with every element in S)

My teacher is horrible so I am pretty lost in the class. I am aware of what the associative property is, but I'm not sure how to go about solving this question when it comes to the binary operation. This is going to be on my exam so I need to know how to solve it.

 

[AdrianZ]

A binary operation is nothing but a two-variable function. It takes two elements of the set G and gives a new element. In other words, a binary operation is a function *: G$\times$G → G and G is a set. Since * is a function on two-variables and its domain is a cartesian product of two sets, the elements in * are ordered pairs like (x,y), we define (x,y) := x*y. If x*y is in G, we say G is closed under the operation *.

Are those definitions clear? Now, your problem asks us to show that H is closed under *. To do that you should take two arbitrary elements of H, like h and h' and show that h*h' is also in H. You don't need to use associativity to solve this problem, the associativity is later needed when you want to show that H is a subgroup of G.

 

[taylor81792]

That helps a lot! Thank you!

 

[Deveno]

A binary operation is nothing but a two-variable function. It takes two elements of the set G and gives a new element. In other words, a binary operation is a function *: G$\times$G → G and G is a set. Since * is a function on two-variables and its domain is a cartesian product of two sets, the elements in * are ordered pairs like (x,y), we define (x,y) := x*y. If x*y is in G, we say G is closed under the operation *.

Are those definitions clear? Now, your problem asks us to show that H is closed under *. To do that you should take two arbitrary elements of H, like h and h' and show that h*h' is also in H. You don't need to use associativity to solve this problem, the associativity is later needed when you want to show that H is a subgroup of G.

(emphasis mine)

yes, you DO. suppose we want to show that h*h' is in H whenever h,h' are. by definition, this means we want to show that:

(h*h')*x = x*(h*h'), for all x in S.

to actually DO this, we might proceed like this:

(h*h')*x = h*(h'*x) <---this is where we need associativity

= h*(x*h') (by definition of H, since h' is in H)

= (h*x)*h' <---associativity used AGAIN

= (x*h)*h' (since h is in H)

= x*(h*h') <---associativity used for a THIRD time.

 

[AdrianZ]

(emphasis mine)

yes, you DO. suppose we want to show that h*h' is in H whenever h,h' are. by definition, this means we want to show that:

(h*h')*x = x*(h*h'), for all x in S.

to actually DO this, we might proceed like this:

(h*h')*x = h*(h'*x) <---this is where we need associativity

= h*(x*h') (by definition of H, since h' is in H)

= (h*x)*h' <---associativity used AGAIN

= (x*h)*h' (since h is in H)

= x*(h*h') <---associativity used for a THIRD time.

indeed. you're right, I didn't do all the steps because the problem looked so straight and simple but you're right.

 

[taylor81792]

Thank you so much